Lectures On Some Fixed Point Theorems Of Functional Analysis

Transcription

1 Lectures On Some Fixed Point Theorems Of Functional Analysis By F.F. Bonsall Tata Institute Of Fundamental Research, Bombay 1962

2 Lectures On Some Fixed Point Theorems Of Functional Analysis By F.F. Bonsall Notes by K.B. Vedak No part of this book may be reproduced in any form by print, microfilm or any other means without written permission from the Tata Institute of Fundamental Research, Colaba, Bombay 5 Tata Institute of Fundamental Research Bombay 1962

3 Introduction These lectures do not constitute a systematic account of fixed point theorems. I have said nothing about these theorems where the interest is essentially topological, and in particular have nowhere introduced the important concept of mapping degree. The lectures have been concerned with the application of a variety of methods to both non-linear (fixed point) problems and linear (eigenvalue) problems in infinite dimensional spaces. A wide choice of techniques is available for linear problems, and I have usually chosen to use those that give something more than existence theorems, or at least a promise of something more. That is, I have been interested not merely in existence theorems, but also in the construction of eigenvectors and eigenvalues. For this reason, I have chosen elementary rather than elegant methods. I would like to draw special attention to the Appendix in which I give the solution due to B. V. Singbal of a problem that I raised in the course of the lectures. I am grateful to Miss K. B. Vedak for preparing these notes and seeing to their publication. Frank F. Bonsall iii

7 Chapter 1 The contraction mapping theorem Given a mapping T of a set E into itself, an element u of E is called a 1 fixed point of the mapping T if Tu=u. Our problem is to find conditions on T and E sufficient to ensure the existence of a fixed point of T in E. We shall also be interested in uniqueness and in procedures for the calculation of fixed points. Definition 1.1. Let E be a nonempty set. A real valued function d defined on E E is called a distance function or metric in E if it satisfies the following conditions i) d(x, y) 0, x, yεe ii) d(x, y)=0 x=y iii) d(x, y)=d(y, x) iv) d(x, z) d(x, y)+d(y, z) A nonempty set with a specified distance function is called a metric space. 1

8 2 The contraction mapping theorem Example. Let X be a set and E denote a set of bounded real valued functions defined on X. Let d be defined on E E by { } d( f, g)=sup f (t) g(t) : tεx, f, gεe. Then d is a metric on E called the uniform metric or uniform distance function. 2 Definition 1.2. A sequence{x n } in a metric space (E, d) is said to converge to an element x of E if lim d(x n, x)=0 n A sequence x n of elements of a metric space (E, d) is called a Cauchy sequence if givenǫ> 0, there exists N such that for p, q N, d(x p, x q )< ǫ. A metric space (E, d) is said to be complete if every Cauchy sequence of its elements converges to an element of E. It is easily verified that each sequence in a metric space converges to at most one on point, and that every convergent sequence is a Cauchy sequence. Example. The space C R [0, 1] of all continuous real valued functions on the closed interval [0, 1] with the uniform distance is a complete metric space. It is not complete in the metric d defined by d ( f, g)= 1 0 f (x) g(x) dx f, gεc R [0, 1]. Definition 1.3. A mapping T of a metric space E into itself is said to satisfy a Lipschitz condition with Lipschitz constant K if d(t x, Ty) Kd(x, y) (x, yεe) If this conditions is satisfied with a Lipschitz constant K such that 0 K < 1 then T is called a contraction mapping. Theorem 1.1 (The contraction mapping theorem). Let T be a contraction mapping of a complete metric space E into itself. Then

12 6 The contraction mapping theorem 8 setting n=0 and x o = u, we have 7 1 d(u r, u) 1 K d(t ru, u)= 1 1 d(t ru, Tu) But d(t r u, Tu) 0 as r. Hence lim d(u r, u)=0 r Example. In the notation of the last example, suppose that y n is a real sequence converging to y o and let T n be the mapping defined on E by (T n φ)(x)=y n + x x o f (t,φ(t))dt Then (T n φ)(x) y o y n y o +mε<δ for n sufficiently large i.e. T n map E into itself for n sufficiently large. Also the mapping T n, T have the same Lipschitz constantεm<1. Obviously for each φεe, lim T n φ=tφ. Hence ifφ n is the unique fixed point of T n (n= n 1, 2,...) then limφ n =φ. In other words, ifφ n is the solution of the n differential equation dy = f (x, y) dx in [x o t, x o +t] with the initial conditionφ n (x o )=y n, thenφ n converges uniformly to the solutionφ withφ(x o )=y o. Remark. The contraction mapping theorem is the simplest of the fixed point theorems that we shall consider. It is concerned with mappings of a complete metric space into itself and in this respect is very general. The theorem is also satisfactory in that the fixed point is always unique and is obtained by an explicit calculation. Its disadvantage is that the condition that the mapping be a contraction is a somewhat severe restriction. In the rest of this chapter we shall obtain certain extension of the contraction mapping theorem in which the conclusion is obtained under modified conditions. Definition 1.4. A mapping T of a metric space E into a metric space E is said to be continuous if for every convergent sequence (x n ) of E, lim n T x n = T( lim n x n ).

15 The contraction mapping theorem 9 11 From (ii) it follows that a sequence (x n ), x n εe is a Cauchy sequence with respect to d ε if and only if it is a Cauchy sequence with respect to d and is convergent with respect to d ε if and only if it converges with respect with respect to d. Hence (E, d) being complete, (E, d ε ) is also a complete metric space. Moreover T is a contraction mapping with respect to d ε. Given x, yεe, and anyε-chain x o,..., x n with x o = x, x n = y, we have d(x i 1, x i )<ε(i=1, 2,...,n), so that d(t x i 1, T x i ) Kd(x i 1, x i )<ε(i=1, 2,...,n) Hence T x o,...,tx n is anε- chain joining T x and T y and d ε (T x, Ty) n d(tc i 1, T x i ) K i=1 n d(x i 1, x i ) i=1 x o,..., x n being an arbitraryε- chain, we have d ε (T x, Ty) Kd ε (x, y) and T has a unique fixed point uεe given by lim d ε(t n x 0, u)=0 for x o εe arbitrary (1) n But in view of the observations made in the beginning of this proof, (1) implies that lim n d(t n x o, u)=0 Example. Let E be a connected compact subset of a domain D in the complex plane. Let f be a complex holomorphic function in D which 12 maps E into itself and satisfies f (z) <1(z,εE). Then there is a unique point z in E with f (z)=z. Since f is continuous in the compact set E, there is a construct K with 0<K< 1 such that f (z) <KεE). For each pointωεe there existsρ ω > 0 such that f (x) is holomorphic in the disc S (ω, 2ρ ω ) of centerωand radius 2ρ ω and satisfies f (z) <K there.

16 10 The contraction mapping theorem E being compact, we can chooseω 1,...,ω n εe such that E is covered by S (ω 1, 2ρ ω1 ),...,S(ω n, 2ρ ωn ) Letε=min{ρω i, i=1, 2,...,n}. If z, z εe and z z <ε then z, z εs (ω i, 2ρ ω1 ) for some i and so f (z) f (z ) = z z f (ω)dω K z z. This proves that Theorem 1.4 is applicable to the mapping z f (z) and we have a unique fixed point. 13 Definition 1.6. A mapping T of a metric space E into itself is said to be contractive if d(t x, Ty)<d(x, y) (x y, x, yεe) and is said to beε-contractive if 0<d(x, y)<ε d(t x, Ty)<d(x, y) Remark. A contractive mapping of a complete metric space into itself need not have a fixed point. e.g. let E ={x/x 1} with the usual distance d(x, y)= x y,let T : E Ebe given by T x= x+ 1 x. Theorem 1.5 (Edelsten). Let T be anε-contractive mapping of a metric space E into itself, and let x o be a point of E such that the sequence (T n x o ) has a subsequence convergent to a point u of E. Then u is a periodic point of T, i.e. there is a positive integer k such that T k u=u Proof. Let (n i ) be a strictly increasing sequence of positive integers such that lim T n i x o = u and let x i = T n i x o. There exists N such that d(x i, u)< i ε/4 for i N. Choose any i Nand let k=n i+1 n i. Then and d(x i+1, T k u)=d(t k x i, T k u) d(x i, u)<ε/4 d(t k u, u) d(t k u, x i+1 )+d(x i+1, u)<ε/2

18 12 The contraction mapping theorem Corollary. If T is a contractive mapping of a metric space E into a compact subset of E, then T has a unique fixed point u in E and u= lim n T n x o where x o is an arbitrary point of E.

19 Chapter 2 Fixed point theorems in normed linear spaces In Chapter 1, we proved fixed point theorems in metric spaces without 16 any algebraic structure. We now consider spaces with a linear structure but non-linear mappings in them. In this chapter we restrict our attention to normed spaces, but our main result will be extended to general locally convex spaces in Chapter 3 Definition 2.1. Let E be a vector space over. A mapping of E into R is called a norm on E if it satisfies the following axioms. i) p(x) 0(x E) ii) p(x)=0 if and only if x=0 iii) p(x+y) p(x)+ p(y) (x, y E). A vector space E with a specified norm on it called a normed space. The norm of an element x E will usually be denoted by x. A normed space is a metric space with the metric d(x, y)= x y (x, yεe) and the corresponding metric topology is called the normed topology. A normed linear space complete in the metric defined by the norm is called a Banach space. We now recall some definitions and well known properties 13

24 18 Fixed point theorems in normed linear spaces Definition 2.4. A subset A of a normed space is said to be bounded if there exists a constant M such that x M (x A). 22 We now state without proof three properties of finite dimensional normed spaces. Lemma 2.2. Every finite dimensional normed space is complete. Lemma 2.3. Every bounded closed subset of a finite dimensional normed space is compact. Lemma 2.4 (Brouwer fixed point theorem). Let K be a non-empty compact convex subset of a finite dimensional normed space, and let T be a continuous mapping of K into itself. Then T has a fixed point in K. The proofs of the first two of these Lemmas are elementary. (Refer to Dunford and Schwartz [14, p ].) The Brouwer fixed point theorem on the other hand is far from trivial. For a proof using some elements of algebraic topology refer to P. Alexandroff and H. Hopf ([1], p ). A proof of a more analytical kind is given by Dunford ans Schwartz ([14], p.467). Theorem 2.2 (Schauder). Let K be a non-empty closed convex subset of a normed space. Let T be a continuous mapping of K into a cumpact subset of K. Then T has fixed point in K. Proof. Let E denote the normed space and let T K A, a compact subset of K. A is contained in a closed convex bounded subset of E. T(B K) T(K) A B so T(B K) is contained in a compact subset of B, K and there is no loss of generality in supposing that K is bounded. If A o is a countable dense subset of the compact metric space A, then the set of all rational linear combinations of elements of A o is a countable dense subset of the closed linear subspace E o spanned by A o and A E 0. Then T(K E 0 ) T(K) A, a compact subset of E 0, and K E 0 is closed and convex. Hence without loss of generality we may assume that K is a bounded closed convex subset of a separable normed space E with a strictly convex norm (Theorem 2.1).

26 20 Fixed point theorems in normed linear spaces Let K be the subset of E consisting of all continuous mappings of [x o ε, x o +ε] into [y o mε, y o + mε]. Then K is a bounded closed convex subset of E. Let T be the mapping defined on K by (Tφ)(x)=y o + x xo f (t,φ(t))dt ( x x o ε) Then T K K. Also since (Tφ)(x) (Tφ)(x ) x f (t, (t))dt m x x (φ K), x T K is an equicontinuous set. Since also T K is bounded, T K is contained in a compact set by the Ascoli - Arzela theorem. Therefore, by Theorem 2.2, T has a fixed pointφin K i.e., φ(x)=y o + x x o f (t,φ(t)) dt ( x x o ε). Thenφis differentiable in [x o ε, x o +ε] and provides a solution y = φ(x) there of the differential equation 25 dy = f (x, y) dx withφ(x o ) = y o. This is Peano s theorem. As a particular case of Schauder s theorem, we have Theorem 2.3. Let K be a non-empty compact convex subset of a normed space, and let T be a continuous mapping of K into itself. Then T has a fixed point in K. Remark. Theorem 2.2 and 2.3 are almost equivalent, in the sense that Theorem 2.2, with the additional hypothesis that K be complete, follows from Theorem 2.3. For, if K is a complete convex set and T K is contained in a compact subset A of K, then the closed convex hull of A is a compact convex subset K o of K, and T K 0 K 0. Definition 2.5. A mapping T which is continuous and maps each bounded set into a compact set is said to be completely continuous.

27 Fixed point theorems in normed linear spaces 21 Theorem 2.4. Let T be a completely continuous mapping of a normed space E into itself and let T E be bounded. Then T has a fixed point. Proof. Let K be the closed convex hull of T E. Then K is bounded and so T K is contained in a compact subset of K. By Theorem 2.2, T has a fixed point in K. The Theorem 2.4 implies Theorem 2.3 is seen as follows. Let K be a compact convex set and let T be continuous mapping of K into itself. There is no loss of generality is supposing that the norm in E is strictly convex. Let P be the metric projection of E onto K, and let 26 T = T P. Then T satisfies the conditions of theorem 24, and so there exists u in E with Tu=u. Since T maps E into K, we have u Kand so Pu=uTu=TRu=u. Lemma 2.5. Let K be a non-empty complete convex subset of a normed space E, let A be a continuous mapping of K into a compact subset of E, and let F be a mapping of K K into K such that (i) F(x, y) F(x, y) k y y (x, y, y K), where k is a constant with 0<k<1, (ii) F(x, y) F(x, y) Ax Ax (x, x, y K). Then there exists a point u in K with F(u, u)=u. Proof. For each fixed x, the mapping y F(x, y) is a contraction mapping of the complete metric space K into itself, and it therefore has a unique fixed point in K which we denote by T x, T x=f(x, T x) (x K). We have T x T x = F(x, T x) F(x, T x ) F(x, T x) F(x, T x) + F(x, T x) F(x, T x ) Ax Ax +k T x T x

28 22 Fixed point theorems in normed linear spaces Therefore T x T x 1 1 k Ax Ax, (1) which shows that Tk is continuous and that T K K is precompact since AK is compact, since K is complete, T K Kis compact. By the Schander theorem, T has fixed point u in K, Tu=u. 27 But then F(u, u)=f(u, Tu)=Tu=u Theorem 2.5 (Kranoselsku ). Let K be a non-empty complete convex subset of a normed space E, let A be a continuous mapping of K into a compact subset of E, let B map K and satisfy a Lipschitz condition Bx Bx k x x (x, x k) with 0<k<1 and let Ax+ By K for all x, y in K. Then there is a point u K with Au+ Bu=u Proof. Take F(x, y) = Ax + By and apply Lemma 2.5. Corollary. Let K be a non-empty complete convex subset of a normed space, let A be a continuous of K into a compact subset of K, let B map K into itself ans satisfy the Lipschitz condition Bx Bx x x (x, x εk), and let 0<α<1. Then there exists a point u K with αau+(1 α)bu=u In general, under the condition of Schauder s theorem, we have no method for the calculation of a fixed point of a mapping. However there is a special case in which this can be done using a method due to Krasnoselsku.

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